If the five-digit number 57xy4 is divisible by 11, how many such numbers are possible?

Odd places (1st, 3rd, 5th): 5 + x + 4. Even places (2nd, 4th): 7 + y.
Difference = (9 + x) − (7 + y) = 2 + x − y, which must be 0 or a multiple of 11.
Case x − y = −2: y = x + 2, so x can run 0 to 7 — 8 numbers.
Case x − y = 9: only x = 9, y = 0 — 1 number.
Total = 9.
Method: always check both the "difference = 0" case and the "difference = 11" case. Stopping at the first gives 8, which is one of the options.

A 25-digit number has identical extreme digits, and all 23 middle digits are identical to each other but different from the extremes. If it is divisible by 11, which value can the end digit not take?

Write the number as a, then 23 copies of b, then a.
Odd places: position 1 (a), positions 3 to 23 (eleven b's), position 25 (a) → 2a + 11b.
Even places: twelve b's → 12b.
Difference = 2a + 11b − 12b = 2a − b, which must be 0 or 11.
2a = b gives a = 1, 2, 3, 4 (with b = 2, 4, 6, 8).
2a − 11 = b gives a = 6, 7, 8, 9 (with b = 1, 3, 5, 7).
So a can be 1, 2, 3, 4, 6, 7, 8 or 9 — never 5.
Note: a = 5 would need b = 10 or b = −1, neither of which is a digit.

Let G be the HCF of 3,780, 3,465 and 3,003, and L be the LCM of 3,465 and 3,003. Find L / 33G.

Factorise: 3780 = 2²·3³·5·7, 3465 = 3²·5·7·11, 3003 = 3·7·11·13.
Common to all three: 3 and 7 at their lowest powers → G = 21.
LCM of 3465 and 3003 takes every prime at its highest power: 3²·5·7·11·13 = 45,045.
L / 33G = 45045 / (33 × 21) = 45045 / 693 = 65.
Note: 5 is absent from 3003, so it cannot appear in the HCF of all three — that is the step where marks are usually lost.

If the units digit of the square of an integer is x, find the sum of all distinct possible values of x.

Square the digits 0 to 9 and note the last digit: 0, 1, 4, 9, 6, 5, 6, 9, 4, 1.
The distinct values are 0, 1, 4, 5, 6, 9.
Sum = 25.
Worth memorising: no perfect square ever ends in 2, 3, 7 or 8. This alone answers several questions a year across different exams.

Find the fraction equivalent to 2.57 (the 7 recurring).

Split it: 2.57 = 2 + 0.57.
For the decimal part, one digit recurs and one does not, so the denominator is 90:
(57 − 5)/90 = 52/90.
Total = 2 + 52/90 = (180 + 52)/90 = 232/90.
Check the bar carefully: if both digits recurred, the answer would be 2 + 57/99 = 255/99, which is not among the options. The position of the bar decides between /90 and /99.

Identify the set of all perfect numbers and twin primes among 1 to 9.

Perfect numbers: 6, since its proper divisors 1 + 2 + 3 = 6. No other number under 10 qualifies.
Twin primes: the pairs (3, 5) and (5, 7) give the members 3, 5 and 7.
Combined set = {3, 5, 6, 7}.
Note: 1 is neither prime nor perfect, and 2 has no twin within the range — both are offered in the distractor sets.